I was doing problem no. $4.4$ from Griffiths Third Edition. I cannot understand one thing related to the solution of the normalization of $Y_2^1$. $$Y_2^1 = -\sqrt{\frac{15}{8\pi}}e^{i\phi}sin\theta cos\theta $$
Where did the $e^{i\phi}$ term go?
I was doing problem no. $4.4$ from Griffiths Third Edition. I cannot understand one thing related to the solution of the normalization of $Y_2^1$. $$Y_2^1 = -\sqrt{\frac{15}{8\pi}}e^{i\phi}sin\theta cos\theta $$
Where did the $e^{i\phi}$ term go?
The squared norm of a vector of a complex Hilbert space is $|v|=+\sqrt{\langle v \vert v\rangle }$, and in this particular Hilbert space the inner product of two vectors is $\langle v|g\rangle =\int d\mu \, v^{*}g$. So in this case, the squared norm would be $\int d\Omega |Y_{2}^{1}|=\int d\Omega\frac{15}{8\pi}\sin^{2}\theta \cos^{2}\theta e^{i\phi} e^{-i\phi}$ where I intentionally did not simplify the last two exponentials for you to know why it "disappears" in the integral.